Question

In an RLC series circuit the value of \(R, L\) and \(C\) is given as \(R = 50\Omega\), \(L = 1.6\,H\) and \(C = 40\mu F\). Find the value inductive reactance \((X_L)\) at resonance.

• A. \(50\Omega\)
• B. \(100\Omega\)
• C. \(200\Omega\)
• D. \(400\Omega\)

Hint:

At resonance in RLC circuit: \[ X_L = X_C \] and \[ \omega = \frac{1}{\sqrt{LC}} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
At resonance in an RLC series circuit, the inductive reactance equals the capacitive reactance ($X_L = X_C$).
The circuit operates at its resonant angular frequency $\omega$.
Step 2: Key Formula or Approach:
The resonant angular frequency is given by $\omega = \frac{1}{\sqrt{LC}}$.
The inductive reactance is $X_L = \omega L$.
Combining these: $X_L = \left(\frac{1}{\sqrt{LC}}\right) L = \sqrt{\frac{L}{C}}$.
Step 3: Detailed Explanation:
Given data:
$L = 1.6 \text{ H}$
$C = 40\mu\text{F} = 40 \times 10^{-6} \text{ F}$
Substitute into the combined formula for $X_L$ at resonance:
\[ X_L = \sqrt{\frac{L}{C}} = \sqrt{\frac{1.6}{40 \times 10^{-6}}} \]
\[ X_L = \sqrt{\frac{1.6}{4 \times 10^{-5}}} = \sqrt{\frac{16 \times 10^{-1}}{4 \times 10^{-5}}} = \sqrt{4 \times 10^4} \]
\[ X_L = 2 \times 10^2 = 200 \ \Omega \]
Step 4: Final Answer:
The value of inductive reactance at resonance is 200 $\Omega$.